WEBVTT 00:00.360 --> 00:04.040 Welcome to this lecture on unsigned integers in the binary system. 00:04.600 --> 00:10.880 A foundational concept in computer organization, especially when working with low level systems. 00:11.280 --> 00:17.600 And today we will take a deep dive into how binary numbers are added and subtracted, how overflow and 00:17.640 --> 00:26.600 underflow behave fixed with binary systems, and how the architectures handles these operations at the 00:26.760 --> 00:27.720 hardware level. 00:28.400 --> 00:33.960 Now, please keep in mind that I will explain everything slowly and clearly, as if I'm writing each 00:33.960 --> 00:35.240 steps on the board. 00:35.760 --> 00:39.640 Whether you are familiar with this topic or completely new, don't worry. 00:39.680 --> 00:42.960 We will build up the understanding step by step. 00:43.560 --> 00:51.160 And in computing, as you have learned so far, we work with three primary numbers or number systems. 00:51.720 --> 00:54.960 We have the decimal which is base ten. 00:55.560 --> 00:58.770 Or let's use the We have decimal base ten. 00:59.410 --> 01:00.370 Base ten. 01:00.970 --> 01:04.210 Now this is the number system most of us use in our daily life. 01:04.210 --> 01:07.050 It includes digits from 0 to 9. 01:07.770 --> 01:11.330 We have binary which is based two. 01:12.130 --> 01:15.250 Now this is the integral language of computers. 01:15.250 --> 01:19.370 It only uses two digits, zeros and ones. 01:19.930 --> 01:23.570 Every calculation inside the computer is done using binary. 01:24.250 --> 01:28.890 And we have the hexadecimal which is base 16. 01:29.610 --> 01:33.730 The system is used to represent binary in a more compact and readable form. 01:34.210 --> 01:49.330 It uses 16 symbols from nine, from 0 to 9, and A to f where A equals ten, b 11, and so on up to 01:49.370 --> 01:52.490 f, which is equals to 15. 01:53.140 --> 01:58.340 Now, even though these numbers systems look different, they can represent the same value. 01:58.580 --> 02:01.300 For example ten in decimal. 02:02.380 --> 02:06.700 Yeah, ten in decimal is not. 02:07.180 --> 02:17.660 Sorry ten in decimal is 1010 in binary and a in hexadecimal. 02:18.020 --> 02:21.220 Now these are just three ways of looking at the same number. 02:21.420 --> 02:28.140 Now you will see these representations frequently in programming, debugging and analyzing data in computer 02:28.140 --> 02:28.740 systems. 02:29.220 --> 02:35.020 Now this table helps you easily translate values between the different number systems. 02:35.020 --> 02:41.860 As you work with more binary and hexadecimal, you will naturally memorize many of this, and you will 02:41.900 --> 02:47.420 often see hexadecimal used when working with memory addresses or machine instructions. 02:47.460 --> 02:52.990 Now, because it's much easier to read and write than long binary sequences like this. 02:53.590 --> 02:58.350 And now what we're going to do is we will add binaries. 02:58.910 --> 03:05.390 Now let's move to the binary addition the process of summing numbers represented in binary. 03:05.590 --> 03:08.990 Now we will use four bit binary numbers in our examples. 03:09.310 --> 03:13.390 That means each number is written using four binary digits. 03:14.030 --> 03:18.510 Now let's represent each number in binary and line them up like a long addition. 03:18.790 --> 03:21.390 So we will add two and four. 03:22.070 --> 03:26.710 So two is what one zero right. 03:26.950 --> 03:34.510 So 0010 and four in binary is 100. 03:34.870 --> 03:37.630 So 0100. 03:38.230 --> 03:40.110 And yeah that's it. 03:40.710 --> 03:45.930 And we also we will also have carry bits carry Pets. 03:46.410 --> 03:57.290 In this case, we have zero carry bits because as you will see, the answer here is zero one again, 03:57.330 --> 04:00.050 one pure one here and pure zero. 04:00.650 --> 04:03.330 And yeah carry bits are 0000. 04:04.050 --> 04:10.090 In this case there are no carry bits generated because the result is less than 16. 04:10.770 --> 04:15.770 Here you can see in the 16th we add one binary. 04:15.770 --> 04:18.570 Here in the eighth we add one binary. 04:18.610 --> 04:21.010 In fourth we add one binary. 04:21.730 --> 04:22.970 This goes on like that. 04:23.650 --> 04:32.370 And here we just basically the generated value result is less than 16 which would require more than 04:32.370 --> 04:33.090 four bits. 04:33.130 --> 04:46.100 Here 16 requires five bits to represent 16 and carry flag is basically zero in a CPU which has which 04:46.100 --> 04:47.780 means no overflow here. 04:48.340 --> 04:50.500 Now this was easy example here. 04:50.980 --> 04:53.980 Now let's do another example. 04:54.620 --> 04:59.940 So what we're going to do we will add four and let's say 14. 05:00.540 --> 05:02.540 And you will see what what will happen here. 05:02.660 --> 05:04.740 So we will have carry bits. 05:05.380 --> 05:10.900 But first we will find four which is 100. 05:11.060 --> 05:19.820 So 0100 and 14 is 1110. 05:20.500 --> 05:27.540 And yeah now let's try adding two larger numbers that will cause an overflow in our four bit space. 05:28.220 --> 05:31.140 And now yeah zero zero. 05:31.140 --> 05:35.940 So there's no carry bit here again zero one. 05:36.700 --> 05:39.870 No zero one is one. 05:40.030 --> 05:40.670 Again. 05:40.670 --> 05:43.150 No carry bit here as well. 05:43.870 --> 05:52.350 But you can see we have two ones which yeah we will add carry bit here and write zero again. 05:53.030 --> 05:54.590 We have the carry bit here. 05:54.910 --> 05:57.670 So you need to translate this bit to here. 05:58.110 --> 06:04.590 And if you translate it to here you will again get zero and still have a carry bit. 06:05.070 --> 06:15.670 Right now if you translate here now you can see that this is basically two which is an incorrect result. 06:16.310 --> 06:23.630 But if we calculate the four and 14 it's basically 18 here right. 06:24.270 --> 06:26.110 Which obviously has five bits. 06:26.630 --> 06:31.430 So I think you are seeing the problem here. 06:31.430 --> 06:31.870 Right. 06:32.470 --> 06:44.680 So 18 in binary is 10010, which needs five bits, but we are only using four bits here, so the extra 06:44.720 --> 06:47.200 bit is dropped somehow. 06:47.600 --> 06:52.640 So this is an what overflow right now. 06:52.680 --> 06:54.720 This is an over. 06:55.400 --> 06:58.840 Now the processor sets the carry flag one. 06:59.560 --> 07:04.320 Now this tells us that the real result doesn't fit in four bits. 07:04.960 --> 07:11.720 And in software, if you want to handle larger numbers, we need to use larger data types like eight 07:11.760 --> 07:18.720 bit or 16 bit integers, or handle the carry manually with logic. 07:19.040 --> 07:26.960 So in this case, what we're going to do is just add another bit here and here. 07:27.000 --> 07:29.200 Put this here right. 07:29.800 --> 07:31.240 And the problem is solved.